Please, still with rotating rigid body problem

[gaborfk]

I am still stuck with this problem. Please, help if you can.
A large, cylindrical roll of tissue paper of initial radius R lies on a long, horizontal surface with the outside end of the paper nailed to the surface. The roll is given a slight shove (V initial = 0) and commences to unroll. Assume the roll has a uniform density and that mechanical energy is conserved in the process. Determine the speed of the center of mass of the roll when its radius has diminished to r = 1.0 mm, assuming R = 6.0 meters.

This is what I got so far.

KE=(1/2)I(Omega)^2

for the cylinder I is :(1/2)MR^2

So mgh=(1/2)I(Omega)^2+(1/2)mv^2

I plugged in I

mgh=(1/2)(1/2)MR^2(Omega)^2+(1/2)mv^2

Then I used (omega)=v/r to get here

mgh=(1/4)MR^2(v/r)^2+(1/2)mv^2 the masses cancel

gh=(1/4)R^2(v/r)^2+(1/2)v^2

Now what? How do I get h? Did I miss anything? Thank you in advance!

 

[arildno]

1. At the start, the total system has only potential energy, MgR, where M is the total mass of the paper.
2. At time "t", you should find that the mass of the remaining rolled-up paper, m, is:
$$m(t)=(\frac{r(t)}{R})^{2}M$$
3. The potential energy of the total system remaining at time "t" is:
$$mgr(t)=Mg\frac{r(t)^{3}}{R^{2}}$$
4. The kinetic energy of the total system at time "t" is:
$$\frac{1}{2}m(t)v^{2}_{G}(t)+\frac{1}{2}(\frac{1}{2}m(t)r(t)^{2})\omega(t)^{2}=\frac{3}{4}(\frac{r(t)}{R})^{2}Mv^{2}_{G}(t)$$
by using the rolling condition to relate the angular velocity to the center of mass of the rolled up paper.
5. Hence, you get:
$$v_{G}(t)=\frac{\sqrt{4g(R^{3}-r^{3}(t))}}{\sqrt{3}r(t)}$$

 

[wais]

Hi there, I can understand your frustration with this problem. It can be tricky to solve at first, but I'll try my best to guide you through it.

Firstly, you're on the right track with using the conservation of mechanical energy to solve this problem. However, there are a few things that need to be clarified.

1. The initial velocity (V initial) is given as 0, which means that the initial kinetic energy (KE) is also 0. So, your equation for mechanical energy conservation should be:

mgh = (1/2)I(omega)^2 + 0

2. The moment of inertia (I) for a cylinder is given by (1/2)MR^2, but in this case, the cylinder is unrolling and its radius is changing. So, the moment of inertia will also change with the radius. We can use the parallel axis theorem to find the moment of inertia at any given radius (r):

I = (1/2)MR^2 + MR^2 = (3/2)MR^2

3. Now, let's take a look at your equation for kinetic energy. You have correctly used the relationship between omega and v, but the mass (m) in your equation should be the mass of the small section of tissue paper that has unrolled, not the entire mass of the cylinder. This means that we need to express the mass in terms of the radius (r) and the initial mass (M) of the cylinder.

m = M(r/R)^2

4. Finally, to solve for the speed (v) when the radius has diminished to r = 1.0 mm, we need to set up the equation with this specific value for r and solve for v.

mgh = (1/4)R^2(omega)^2 + (1/2)m(r/R)^2v^2

Substituting in the values for m, I, and h, we get:

(1/4)MR^2gh = (1/4)(3/2)MR^2(omega)^2 + (1/2)M(r/R)^2v^2

Canceling out the common terms and solving for v, we get:

v = sqrt(2gh(R^2-r^2)/3R^2)

Plugging in the given values of g, h, and R, we can calculate the